Complex Numbers in AS Further Maths — AS AQA 进阶数学：复数完全指南

1. What Is a Complex Number? Breaking Down the Imaginary Unit i | 什么是复数？拆解虚数单位 i

复数（Complex Numbers）是 AS 进阶数学中最具革命性的概念之一。在实数系统中，负数的平方根没有定义 – 比如 √(-1) 在实数轴上找不到对应点。数学家引入虚数单位 i（定义 i² = -1），将数的世界从一维实数轴拓展到二维复平面。任何复数都可以写成 z = a + bi 的形式，其中 a 是实部（Real Part），b 是虚部（Imaginary Part），a 和 b 都是实数。

Complex numbers are one of the most revolutionary concepts in AS Further Mathematics. In the real number system, the square root of a negative number is undefined – for example, √(-1) has no corresponding point on the real axis. Mathematicians introduced the imaginary unit i (defined such that i² = -1), expanding the number world from a one-dimensional real line to a two-dimensional complex plane. Any complex number can be written as z = a + bi, where a is the real part, b is the imaginary part, and both a and b are real numbers.

理解复数的关键在于认识到 i 不是一个”虚构”的数，而是一个旋转算子。在复平面（Argand Diagram）上，乘以 i 相当于逆时针旋转 90°。这一几何直观解释了为什么 i² = -1：旋转 180° 正好指向相反方向。AQA 考试中，你不需要证明这一点，但掌握几何含义能帮助你快速验证代数运算结果。

The key to understanding complex numbers is recognising that i is not a “fake” number – it is a rotation operator. On the complex plane (Argand Diagram), multiplying by i is equivalent to a 90° counterclockwise rotation. This geometric intuition explains why i² = -1: rotating 180° points in the exact opposite direction. In AQA exams, you don’t need to prove this, but grasping the geometric meaning helps you quickly verify algebraic results.

2. Adding and Subtracting Complex Numbers: The Component-Wise Rule | 复数的加减法：分量分别运算规则

复数的加法和减法遵循分量分别运算原则 – 实部与实部运算，虚部与虚部运算。若 z₁ = a + bi 且 z₂ = c + di，则 z₁ + z₂ = (a + c) + (b + d)i，z₁ – z₂ = (a – c) + (b – d)i。这一规则之所以成立，是因为在复平面上，复数加法对应向量加法 – 将两个复数的实部和虚部分别叠加，等价于将它们在 Argand Diagram 上首尾相连。

Complex number addition and subtraction follow a component-wise rule – real parts combine with real parts, imaginary parts with imaginary parts. If z₁ = a + bi and z₂ = c + di, then z₁ + z₂ = (a + c) + (b + d)i, and z₁ – z₂ = (a – c) + (b – d)i. This rule holds because on the complex plane, complex addition corresponds to vector addition – adding the real and imaginary components separately is equivalent to connecting the two numbers tip-to-tail on the Argand Diagram.

AQA 考试中常见的陷阱：当虚部为负时，学生容易在加减法中遗漏负号。例如计算 (3 – 4i) + (-2 + 7i)，许多学生会把 -4i + 7i 算成 -11i，正确结果应该是 +3i。建议在草稿纸上明确写出每一项的符号，用括号包裹每个复数再进行运算。

A common pitfall in AQA exams: when the imaginary part is negative, students often drop the minus sign during addition or subtraction. For example, when calculating (3 – 4i) + (-2 + 7i), many students compute -4i + 7i as -11i – the correct result is +3i. Always write out each term with its sign explicitly, and bracket each complex number before performing the operation.

3. Multiplying Complex Numbers: FOIL Method and the i² = -1 Simplification | 复数乘法：FOIL 展开法与 i² = -1 化简

复数乘法看起来复杂，但只需记住一个核心步骤：像展开二项式一样使用 FOIL 法则（先乘首项、外项、内项、末项），然后将所有 i² 替换为 -1。例如 (2 + 3i)(4 – i) = 8 – 2i + 12i – 3i² = 8 + 10i – 3(-1) = 11 + 10i。关键在于最后一步 – 合并实部与虚部之前，必须将 i² 替换为 -1。

Complex multiplication looks daunting but only requires one core step: expand using the FOIL method (First, Outer, Inner, Last) as if multiplying binomials, then replace every i² with -1. For example, (2 + 3i)(4 – i) = 8 – 2i + 12i – 3i² = 8 + 10i – 3(-1) = 11 + 10i. The critical step is replacing i² with -1 before combining real and imaginary parts.

对于形如 (a + bi)(a – bi) 的共轭复数乘积，结果总是实数 a² + b²。这是因为 (a + bi)(a – bi) = a² – (bi)² = a² – b²i² = a² + b²。这一性质在复数除法中至关重要 – 分母有理化的关键就是乘以分母的共轭复数。

For conjugate complex products of the form (a + bi)(a – bi), the result is always the real number a² + b². This is because (a + bi)(a – bi) = a² – (bi)² = a² – b²i² = a² + b². This property is essential for complex division – the key to rationalising the denominator is multiplying by the denominator’s complex conjugate.

4. The Complex Conjugate: Definition, Notation, and Why It Matters | 共轭复数：定义、记法及其重要性

复数 z = a + bi 的共轭复数记为 z*（或写作 z̄），定义为 z* = a – bi – 将虚部符号取反即可。在 Argand Diagram 上，z 与 z* 关于实轴对称。共轭复数的重要性体现在三个方面：(1) 复数除法的核心工具；(2) 二次方程根的性质 – 若系数为实数，复根必成对出现，且互为共轭；(3) 求复数的模 – z × z* = |z|² = a² + b²。

The complex conjugate of z = a + bi, denoted z* (or z̄), is defined as z* = a – bi – simply negate the imaginary part. On the Argand Diagram, z and z* are symmetric about the real axis. The complex conjugate is important for three reasons: (1) it is the core tool for complex division; (2) it governs the nature of quadratic roots – if the coefficients are real, complex roots always appear in conjugate pairs; and (3) it gives the modulus – z × z* = |z|² = a² + b².

AQA 考试每年都会有题目要求”写出 z = … 的共轭复数”，这是送分题，但务必注意虚部符号。例如 z = -3 + 5i 的共轭是 -3 – 5i（不是 3 – 5i），虚部符号取反即可，实部保持不变。

AQA exams consistently feature a question asking you to “write down the conjugate of z = …” – this is a guaranteed mark-earner, but be careful with the sign. For example, the conjugate of z = -3 + 5i is -3 – 5i (not 3 – 5i); only the imaginary part changes sign, the real part stays the same.

5. Dividing Complex Numbers: Multiplying by the Conjugate Denominator | 复数除法：乘以分母的共轭复数

复数除法的核心策略是将分母”实数化” – 分子分母同时乘以分母的共轭复数。例如计算 (3 + 2i) ÷ (1 – i)：分子分母同乘 (1 + i)，得到 [(3 + 2i)(1 + i)] / [(1 – i)(1 + i)] = (3 + 3i + 2i + 2i²) / (1² + 1²) = (3 + 5i – 2) / 2 = (1 + 5i) / 2 = 0.5 + 2.5i。最终必须写成 a + bi 的标准形式。

The core strategy for complex division is “real-ising” the denominator – multiply both numerator and denominator by the denominator’s complex conjugate. For example, to compute (3 + 2i) ÷ (1 – i): multiply top and bottom by (1 + i), giving [(3 + 2i)(1 + i)] / [(1 – i)(1 + i)] = (3 + 3i + 2i + 2i²) / (1² + 1²) = (3 + 5i – 2) / 2 = (1 + 5i) / 2 = 0.5 + 2.5i. The final answer must be expressed in standard a + bi form.

考试中常见的扣分点：除法完成后忘记将结果整理成 a + bi 形式。如果答案写成 (1 + 5i)/2，AQA 评分标准通常只给方法分，最终答案分要求写成 0.5 + 2.5i。另外，当分母为纯虚数（如 2i）时，可以直接处理而不需要共轭：a/(bi) = (a × -i) / (b) = -ai/b，这比乘以共轭更快。

A common mark-loser in exams: forgetting to express the final result in a + bi form after division. If the answer is left as (1 + 5i)/2, the AQA mark scheme typically awards method marks only – the final answer mark requires 0.5 + 2.5i. Also, when the denominator is purely imaginary (e.g., 2i), you can handle it directly without the conjugate: a/(bi) = (a × -i) / b = -ai/b, which is faster than multiplying by the conjugate.

6. The Argand Diagram: Visualising Complex Numbers on a Plane | Argand 图：在平面上可视化复数

Argand Diagram（阿甘图）将复数映射到二维平面上：横轴为实轴（Real Axis），纵轴为虚轴（Imaginary Axis）。复数 z = a + bi 对应坐标为 (a, b) 的点。这一表示方式将代数问题转化为几何问题 – 复数的加法是向量加法，模长是点到原点的距离，辐角是点与正实轴的夹角。AQA 考试要求你能够：(1) 在 Argand Diagram 上标出给定复数；(2) 解释复数运算的几何含义；(3) 用模长和辐角表示复数（极坐标形式）。

The Argand Diagram maps complex numbers onto a two-dimensional plane: the horizontal axis is the real axis, and the vertical axis is the imaginary axis. The complex number z = a + bi corresponds to the point (a, b). This representation turns algebraic problems into geometric ones – complex addition is vector addition, the modulus is the distance from the point to the origin, and the argument is the angle the point makes with the positive real axis. The AQA exam expects you to: (1) plot given complex numbers on an Argand Diagram; (2) explain the geometric meaning of complex operations; (3) express complex numbers in modulus-argument (polar) form.

一道典型的 AQA AS 考题：在 Argand Diagram 上标出 z₁ = 3 + 4i, z₂ = -1 + 2i, 以及 z₁ + z₂，并说明它们构成的几何关系。答案是这三个点形成一个平行四边形 – z₁ 和 z₂ 是从原点出发的两条边，z₁ + z₂ 是对角线。这完美展示了复数加法与向量加法的等价关系。

A typical AQA AS exam question: plot z₁ = 3 + 4i, z₂ = -1 + 2i, and z₁ + z₂ on an Argand Diagram, and describe the geometric relationship they form. The answer: these three points form a parallelogram – z₁ and z₂ are two sides from the origin, and z₁ + z₂ is the diagonal. This beautifully demonstrates the equivalence between complex addition and vector addition.

7. Modulus and Argument: The Distance and Direction of a Complex Number | 模长与辐角：复数的距离与方向

复数的模（Modulus）|z| = √(a² + b²)，表示复平面上点到原点的距离。辐角（Argument）arg(z) 是复平面上点与正实轴的夹角，通常以弧度表示，范围在 -π 到 π 之间（主值范围）。在 AS 进阶数学中，你需要能够：给定 a + bi 形式，求模和辐角；给定模和辐角，还原 a + bi 形式；以及理解 z × z* = |z|² 这一关键恒等式。

The modulus of a complex number, |z| = √(a² + b²), represents the distance from the point to the origin on the complex plane. The argument, arg(z), is the angle the point makes with the positive real axis, usually expressed in radians and within the range -π to π (the principal value). In AS Further Mathematics, you need to be able to: find the modulus and argument from a + bi form; reconstruct a + bi form from modulus and argument; and understand the crucial identity z × z* = |z|².

求辐角时最常见的错误是使用错误的反正切分支。例如 z = -1 + i，tan⁻¹(1/(-1)) = tan⁻¹(-1) = -π/4，但该点位于第二象限，正确辐角应该是 π – π/4 = 3π/4。必须根据 a 和 b 的正负号判断象限来调整结果：第一象限 arg = tan⁻¹(b/a)；第二象限 arg = π – tan⁻¹(|b/a|)；第三象限 arg = -π + tan⁻¹(|b/a|)；第四象限 arg = -tan⁻¹(|b/a|)。

The most common error when finding the argument is using the wrong arctangent branch. For example, with z = -1 + i, tan⁻¹(1/(-1)) = tan⁻¹(-1) = -π/4, but the point lies in the second quadrant – the correct argument is π – π/4 = 3π/4. You must adjust the result based on the signs of a and b: First quadrant: arg = tan⁻¹(b/a); Second quadrant: arg = π – tan⁻¹(|b/a|); Third quadrant: arg = -π + tan⁻¹(|b/a|); Fourth quadrant: arg = -tan⁻¹(|b/a|).

8. Solving Quadratic Equations with Complex Roots: When the Discriminant Is Negative | 解有复数根的二次方程：判别式为负时

在 AS 进阶数学中，二次方程 ax² + bx + c = 0 的判别式 Δ = b² – 4ac 决定根的性质。当 Δ < 0 时，方程没有实数根，但有两个共轭复根。使用求根公式 x = [-b ± √(b² - 4ac)] / (2a)，其中 √(b² - 4ac) = √(4ac - b²) × i。例如 x² + 4x + 13 = 0：Δ = 16 - 52 = -36，x = [-4 ± √(-36)] / 2 = [-4 ± 6i] / 2 = -2 ± 3i。两个根 -2 + 3i 和 -2 - 3i 互为共轭。

In AS Further Mathematics, the discriminant Δ = b² – 4ac of a quadratic equation ax² + bx + c = 0 determines the nature of its roots. When Δ < 0, the equation has no real roots but instead has a pair of complex conjugate roots. Use the quadratic formula x = [-b ± √(b² - 4ac)] / (2a), where √(b² - 4ac) = √(4ac - b²) × i. For example, with x² + 4x + 13 = 0: Δ = 16 - 52 = -36, so x = [-4 ± √(-36)] / 2 = [-4 ± 6i] / 2 = -2 ± 3i. The two roots -2 + 3i and -2 - 3i are complex conjugates of each other.

这一性质可以推广到任何实系数多项式方程：复根总是成对出现且互为共轭。AQA 考试经常考”已知方程有一个复根，求另一个根以及未知系数”的题型。例如已知 3 + 2i 是 x² + px + q = 0 的一个根，求 p 和 q。解：另一根为 3 – 2i，使用韦达定理，两根之和 = -p = 6，所以 p = -6；两根之积 = q = (3+2i)(3-2i) = 9 + 4 = 13。

This property extends to any polynomial equation with real coefficients: complex roots always appear in conjugate pairs. AQA exams frequently feature questions like: “Given that one root of the equation is complex, find the other root and the unknown coefficients.” For example, given that 3 + 2i is a root of x² + px + q = 0, find p and q. Solution: the other root is 3 – 2i. Using Vieta’s formulas, sum of roots = -p = 6, so p = -6; product of roots = q = (3+2i)(3-2i) = 9 + 4 = 13.

9. Modulus-Argument Form: Writing z = r(cos θ + i sin θ) | 模-辐角形式：z = r(cos θ + i sin θ) 的写法

在 AS 进阶数学中，复数可以用模-辐角形式（Polar Form 极坐标形式）表示：z = r(cos θ + i sin θ)，其中 r = |z| 是模，θ = arg(z) 是辐角。这一形式将复数的代数和几何表示完美统一。从 a + bi 转化为极坐标形式：r = √(a² + b²)，θ = arctan(b/a)（需根据象限调整）。逆转化：a = r cos θ，b = r sin θ。

In AS Further Mathematics, complex numbers can be expressed in modulus-argument form (polar form): z = r(cos θ + i sin θ), where r = |z| is the modulus and θ = arg(z) is the argument. This form elegantly unifies the algebraic and geometric representations of complex numbers. Converting from a + bi to polar form: r = √(a² + b²), θ = arctan(b/a) (adjusted by quadrant). Reverse conversion: a = r cos θ, b = r sin θ.

模-辐角形式在复数乘除法中展现出巨大优势：两个复数相乘，模相乘、辐角相加 – |z₁z₂| = |z₁| × |z₂|，arg(z₁z₂) = arg(z₁) + arg(z₂)。除法类似：模相除，辐角相减。这一性质在 AQA 考试中的几何应用题中频繁出现，例如”描述乘以 (1 + i) 对复平面上任意点的影响” – 答案是模变为原来的 √2 倍，辐角增加 π/4（即旋转 45° 并缩放 1.414 倍）。

The modulus-argument form reveals a powerful advantage in complex multiplication and division: when multiplying two complex numbers, moduli multiply and arguments add – |z₁z₂| = |z₁| × |z₂|, arg(z₁z₂) = arg(z₁) + arg(z₂). Division works similarly: moduli divide, arguments subtract. This property appears frequently in AQA geometry application questions, e.g., “Describe the effect of multiplying any point on the complex plane by (1 + i).” The answer: the modulus is scaled by √2, and the argument increases by π/4 (a 45-degree rotation and 1.414× scaling).

10. Complex Roots of Unity: Solving zⁿ = 1 and Cubic Roots in Particular | 单位根：解 zⁿ = 1 及其三次方根

方程 zⁿ = 1 的解称为 n 次单位根（Roots of Unity），共有 n 个解，均匀分布在复平面上的单位圆上。对于 AS 进阶数学，最常见的是三次单位根 z³ = 1。除了显而易见的 z = 1，另外两个根是 z = -½ ± (√3/2)i，分别记为 ω 和 ω²。这三个根满足 1 + ω + ω² = 0 和 ω³ = 1。AQA 考试有时会考利用 ω 的性质化简复杂表达式。

The solutions to zⁿ = 1 are called the nth roots of unity. There are exactly n solutions, equally spaced around the unit circle on the complex plane. For AS Further Mathematics, the most common case is the cube roots of unity from z³ = 1. Besides the obvious z = 1, the other two roots are z = -½ ± (√3/2)i, conventionally denoted ω and ω². These three roots satisfy 1 + ω + ω² = 0 and ω³ = 1. AQA exams sometimes test simplification of complex expressions using the properties of ω.

更大次数的单位根（如 4 次、5 次）也可能出现在 AS 试卷中，解题思路相同：使用极坐标形式 z = cos(2kπ/n) + i sin(2kπ/n)，其中 k = 0, 1, 2, …, n-1。关键是将 zⁿ = 1 改写为 zⁿ = cos(2kπ) + i sin(2kπ)，然后用 De Moivre 定理提取 n 次根。

Higher-order roots of unity (such as 4th or 5th roots) may also appear in AS papers. The solution approach is the same: use polar form z = cos(2kπ/n) + i sin(2kπ/n), where k = 0, 1, 2, …, n-1. The key is rewriting zⁿ = 1 as zⁿ = cos(2kπ) + i sin(2kπ), then applying De Moivre’s Theorem to extract the nth root.

11. AQA Exam Technique: Maximising Marks on Complex Number Questions | AQA 考试技巧：复数题型如何最大化得分

AQA AS 进阶数学中，复数通常出现在纯数部分（Paper 1），占 8-12 分（约 10%-15% 的总分）。考试题型包括：基础运算（加减乘除共轭）、解二次方程、Argand Diagram 作图与几何意义、模和辐角的计算、以及综合应用题。以下策略可以帮助你最大化得分：(1) 每一步都写出清晰的过程 – 乘法展示 FOIL 展开，除法展示分母共轭操作；(2) 最终答案永远写成 a + bi 或 r(cos θ + i sin θ) 形式；(3) 画图验证 – 在 Argand Diagram 上检查你的答案是否在期望的象限。

In AQA AS Further Mathematics, complex numbers typically appear in the Pure section (Paper 1), carrying 8-12 marks (approximately 10-15% of the total). Exam question types include: basic operations (addition, subtraction, multiplication, division, conjugates), solving quadratic equations, Argand Diagram plotting and geometric interpretation, calculation of modulus and argument, and integrated application questions. These strategies will maximise your marks: (1) show clear working for every step – display FOIL expansion for multiplication and conjugate operations for division; (2) always present the final answer in a + bi or r(cos θ + i sin θ) form; (3) sketch and verify – check on an Argand Diagram that your answer lies in the expected quadrant.

最容易丢分的地方往往不是概念理解，而是细节处理：(a) 忘记将结果写成标准形式；(b) 辐角计算时未考虑象限；(c) 虚部为负时的符号错误；(d) 解方程时只给了一个根，忘写共轭根。养成检查习惯：模是否为正？共轭根是否成对？除法的分母是否已实数化？

The marks most commonly lost are not from conceptual misunderstanding but from detail handling: (a) forgetting to express the result in standard form; (b) failing to consider the quadrant when calculating the argument; (c) sign errors when the imaginary part is negative; (d) solving an equation but listing only one root, forgetting the conjugate root. Develop checking habits: is the modulus positive? Do the conjugate roots appear in pairs? Has the denominator been real-ised in division?

12. Geometric Locus Problems: Describing Sets of Points on the Argand Diagram | 几何轨迹问题：描述 Argand 图上的点集

AQA AS 进阶数学中，轨迹（Locus）问题是复数章节的高频考点。典型的题型包括：(1) |z – a| = r：以 a 为圆心、r 为半径的圆；(2) |z – a| = |z – b|：点 a 和点 b 的垂直平分线；(3) arg(z – a) = θ：以 a 为起点、与正实轴成 θ 角的半射线。解答这类题的关键是将代数不等式翻译成几何图形，然后在 Argand Diagram 上标注。

In AQA AS Further Mathematics, locus problems are a high-frequency topic in the complex numbers chapter. Typical question types include: (1) |z – a| = r: a circle centred at a with radius r; (2) |z – a| = |z – b|: the perpendicular bisector of the segment joining points a and b; (3) arg(z – a) = θ: a half-line starting at a, making an angle θ with the positive real axis. The key to answering these questions is translating the algebraic inequality into a geometric shape, then annotating it on the Argand Diagram.

组合不等式是难度升级的考点。例如”在 Argand Diagram 上画出满足 |z – 2| < 3 且 arg(z) > π/4 的点的区域”。|z – 2| < 3 表示以 (2, 0) 为圆心、半径为 3 的开圆盘；arg(z) > π/4 表示从原点出发、与正实轴成 45° 角的射线以上的区域。两个条件的交集是一个扇形。画图时必须明确标注边界是否包含（虚线表示不包含，实线表示包含）。

Combined inequalities represent a higher difficulty level. For example: “On an Argand Diagram, shade the region of points satisfying |z – 2| < 3 and arg(z) > π/4.” The condition |z – 2| < 3 describes an open disc centred at (2, 0) with radius 3; arg(z) > π/4 describes the region above the ray from the origin at 45° to the positive real axis. The intersection of these two conditions yields a sector. When sketching, you must clearly indicate whether boundaries are included (dashed line for excluded, solid line for included).

13. De Moivre’s Theorem: Powers and Roots Made Simple | 棣莫弗定理：幂与根运算的简化

棣莫弗定理（De Moivre’s Theorem）是处理复数乘方和最简根式的利器，但在 AS 进阶数学中仅需掌握基础应用。定理陈述：对于任意整数 n，[r(cos θ + i sin θ)]ⁿ = rⁿ(cos nθ + i sin nθ)。也就是说，对一个复数取 n 次方，模变成原来的 n 次方，辐角乘以 n。这一性质使得计算如 (1 + i)⁸ 这样的高次幂变得极其简单 – 先转化为极坐标形式 1 + i = √2(cos π/4 + i sin π/4)，然后 (√2)⁸(cos 2π + i sin 2π) = 16(1 + 0i) = 16。

De Moivre’s Theorem is a powerful tool for handling complex powers and roots, though in AS Further Mathematics only the basic applications are required. The theorem states: for any integer n, [r(cos θ + i sin θ)]ⁿ = rⁿ(cos nθ + i sin nθ). In other words, when raising a complex number to the nth power, the modulus is raised to the nth power and the argument is multiplied by n. This property makes calculating high powers like (1 + i)⁸ extremely simple – first convert to polar form: 1 + i = √2(cos π/4 + i sin π/4), then (√2)⁸(cos 2π + i sin 2π) = 16(1 + 0i) = 16.

De Moivre 定理的另一重要应用是求解 zⁿ = w 形式的方程。将 w 转化为极坐标形式，然后 z 的第 k 个根为 r^(1/n)[cos((θ + 2kπ)/n) + i sin((θ + 2kπ)/n)]，其中 k = 0, 1, …, n-1。例如 z⁴ = 16i：|16i| = 16，辐角为 π/2，四个根分别对应 k = 0, 1, 2, 3，均匀分布在以原点为圆心、2 为半径的圆上，相邻根之间的夹角为 90°。

Another important application of De Moivre’s Theorem is solving equations of the form zⁿ = w. Convert w to polar form, then the kth root of z is r^(1/n)[cos((θ + 2kπ)/n) + i sin((θ + 2kπ)/n)], where k = 0, 1, …, n-1. For example, z⁴ = 16i: |16i| = 16 with argument π/2. The four roots correspond to k = 0, 1, 2, 3, evenly spaced around a circle of radius 2 centred at the origin, with an angular separation of 90° between consecutive roots.

14. AQA Exam Practice: Full Worked Solution for a Typical Complex Numbers Question | AQA 真题演练：一道典型复数考题的完整解答

以下是一道典型的 AQA AS 进阶数学复数综合题，涵盖了本章节的核心技能。题目：已知 z = 2 – 3i，(a) 求 z* 和 |z|；(b) 计算 z² 并以 a + bi 形式表示；(c) 求 (1 + 2i) / z 的结果；(d) 在 Argand Diagram 上标出 z, z*, z²，并说明它们的几何关系。

Below is a typical AQA AS Further Mathematics integrated complex numbers question, covering the core skills of this chapter. Question: Given z = 2 – 3i, (a) find z* and |z|; (b) compute z² and express it in a + bi form; (c) find (1 + 2i) / z; (d) plot z, z*, and z² on an Argand Diagram, and describe their geometric relationship.

解答 (a)：共轭复数 z* = 2 + 3i。模长 |z| = √(2² + (-3)²) = √(4 + 9) = √13。(b)：z² = (2 – 3i)² = 4 – 12i + 9i² = 4 – 12i – 9 = -5 – 12i。(c)：(1 + 2i)/(2 – 3i)，分子分母同乘 (2 + 3i)：[(1+2i)(2+3i)] / [(2-3i)(2+3i)] = (2 + 3i + 4i + 6i²) / (4 + 9) = (2 + 7i – 6) / 13 = (-4 + 7i) / 13 = -4/13 + (7/13)i。最终答案可以保留分数形式：-4/13 + (7/13)i。(d)：z = (2, -3) 位于第四象限，z* = (2, 3) 位于第一象限 – 两者关于实轴对称。z² = (-5, -12) 位于第三象限，|z²| = 13 = |z|²，arg(z²) = 2 × arg(z)，展示了模平方、辐角加倍的几何关系。

Solution (a): The complex conjugate z* = 2 + 3i. The modulus |z| = √(2² + (-3)²) = √(4 + 9) = √13. (b): z² = (2 – 3i)² = 4 – 12i + 9i² = 4 – 12i – 9 = -5 – 12i. (c): (1 + 2i)/(2 – 3i), multiply numerator and denominator by (2 + 3i): [(1+2i)(2+3i)] / [(2-3i)(2+3i)] = (2 + 3i + 4i + 6i²) / (4 + 9) = (2 + 7i – 6) / 13 = (-4 + 7i) / 13 = -4/13 + (7/13)i. The final answer can be left in fraction form: -4/13 + (7/13)i. (d): z = (2, -3) lies in the fourth quadrant, z* = (2, 3) lies in the first quadrant – they are symmetric about the real axis. z² = (-5, -12) lies in the third quadrant, |z²| = 13 = |z|², arg(z²) = 2 × arg(z), demonstrating the geometric relationship: modulus squared, argument doubled.

Summary | 总结

复数是 AS 进阶数学 AQA 课程中连接代数与几何的桥梁。掌握复数的四种基本运算（加减乘除）、理解共轭和模-辐角的双重表示、熟练运用 Argand Diagram 进行几何分析，是应对 AQA 考试的三大核心能力。记住：i 不是”虚幻的”，它是旋转操作 – 每乘一次 i，就在复平面上逆时针转 90°。从解二次方程到求单位根，复数系统为看似”无解”的问题提供了优雅的答案。

Complex numbers are the bridge connecting algebra and geometry in the AS Further Mathematics AQA curriculum. Mastering the four basic operations (addition, subtraction, multiplication, division), understanding the dual representation of conjugates and modulus-argument form, and becoming proficient in geometric analysis using the Argand Diagram are the three core competencies for tackling AQA exams. Remember: i is not “imaginary” – it is a rotation operator; every multiplication by i rotates a point 90° counterclockwise on the complex plane. From solving quadratics to finding roots of unity, the complex number system provides elegant answers to problems that seem “unsolvable.”

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